Although it is not evident at first, factor groups correspond exactly to homomorphic images,
and we can use factor groups to study homomorphisms. We already know that with every
group homomorphism*ϕ*:*G→H*we can associate a normal subgroup of*G, kerϕ. The con-*
verse is also true; that is, every normal subgroup of a group*G*gives rise to homomorphism
of groups.

Let*H* be a normal subgroup of*G. Define the natural*or

**canonical homomorphism***ϕ*:

*G→G*/

*H*

by

*ϕ*(*g*) =*gH.*
This is indeed a homomorphism, since

*ϕ*(*g*1*g*2) =*g*1*g*2*H* =*g*1*Hg*2*H*=*ϕ*(*g*1)*ϕ*(*g*2)*.*

The kernel of this homomorphism is*H. The following theorems describe the relationships*
between group homomorphisms, normal subgroups, and factor groups.

**Theorem 11.10 First Isomorphism Theorem.** *Ifψ*:*G→H* *is a group homomorphism*
*withK* =ker*ψ, thenK* *is normal inG. Letϕ*:*G→G*/*K* *be the canonical homomorphism.*
*Then there exists a unique isomorphism* *η*:*G*/*K* *→ψ*(*G*) *such that* *ψ*=*ηϕ.*

Proof. We already know that *K* is normal in *G. Defineη* :*G*/*K* *→* *ψ*(*G*) by *η*(*gK*) =

*ψ*(*g*). We first show that *η* is a well-defined map. If *g*1*K* = *g*2*K, then for some* *k* *∈* *K,*

*g*1*k*=*g*2; consequently,

11.2. THE ISOMORPHISM THEOREMS 127
Thus,*η*does not depend on the choice of coset representatives and the map*η*:*G*/*K→ψ*(*G*)

is uniquely defined since*ψ*=*ηϕ. We must also show thatη* is a homomorphism, but
*η*(*g*1*Kg*2*K*) =*η*(*g*1*g*2*K*)

=*ψ*(*g*1*g*2)

=*ψ*(*g*1)*ψ*(*g*2)

=*η*(*g*1*K*)*η*(*g*2*K*)*.*

Clearly, *η* is onto *ψ*(*G*). To show that *η* is one-to-one, suppose that *η*(*g*1*K*) = *η*(*g*2*K*).

Then *ψ*(*g*1) =*ψ*(*g*2). This implies that*ψ*(*g*1*−*1*g*2) =*e, org−*11*g*2 is in the kernel of*ψ; hence,*

*g−*_{1}1*g*2*K* =*K; that is,* *g*1*K*=*g*2*K.*

Mathematicians often use diagrams called * commutative diagrams* to describe such
theorems. The following diagram “commutes” since

*ψ*=

*ηϕ.*

*ψ*

*ϕ*

*η*

*G*

*H*

*G*/*K*

**Example 11.11.** Let*G* be a cyclic group with generator *g. Define a mapϕ* :Z*→* *G* by
*n7→gn*. This map is a surjective homomorphism since

*ϕ*(*m*+*n*) =*gm*+*n*=*gmgn*=*ϕ*(*m*)*ϕ*(*n*)*.*

Clearly *ϕ*is onto. If * _{|}g|*=

*m, thengm*=

*e. Hence, kerϕ*=

*m*Zand

_{Z}/ker

*ϕ*=Z/

*m*Z

*∼*=

*G.*

On the other hand, if the order of *g* is infinite, then ker*ϕ*= 0and *ϕ*is an isomorphism of
*G*and Z. Hence, two cyclic groups are isomorphic exactly when they have the same order.
Up to isomorphism, the only cyclic groups are_{Z} and_{Z}*n*.

**Theorem 11.12 Second Isomorphism Theorem.** *Let* *H* *be a subgroup of a group* *G*
*(not necessarily normal in* *G) and* *N* *a normal subgroup of* *G. Then* *HN* *is a subgroup of*
*G,* *H∩N* *is a normal subgroup ofH, and*

*H*/*H∩N* *∼*=*HN*/*N.*

Proof. We will first show that*HN* =*{hn*:*h∈H, n∈N}*is a subgroup of*G. Suppose*
that*h*1*n*1*, h*2*n*2*∈HN. Since* *N* is normal, (*h*2)*−*1*n*1*h*2 *∈N*. So

(*h*1*n*1)(*h*2*n*2) =*h*1*h*2((*h*2)*−*1*n*1*h*2)*n*2

is in *HN. The inverse of* *hn∈HN* is in *HN* since

Next, we prove that *H* *∩N* is normal in *H. Let* *h* *∈* *H* and *n* *∈* *H* *∩* *N*. Then
*h−*1_{nh}_{∈}_{H}_{since each element is in}* _{H. Also,}_{h}−*1

_{nh}_{∈}_{N}_{since}

_{N}_{is normal in}

_{G; therefore,}*h−*1*nh∈H∩N*.

Now define a map*ϕ*from*H* to*HN*/*N* by *h7→hN. The mapϕ*is onto, since any coset
*hnN* =*hN* is the image of*h* in*H. We also know thatϕ*is a homomorphism because

*ϕ*(*hh′*) =*hh′N* =*hN h′N* =*ϕ*(*h*)*ϕ*(*h′*)*.*

By the First Isomorphism Theorem, the image of *ϕ*is isomorphic to*H*/ker*ϕ; that is,*
*HN*/*N* =*ϕ*(*H*)*∼*=*H*/ker*ϕ.*

Since

ker*ϕ*=*{h∈H*:*h∈N}*=*H∩N,*
*HN*/*N* =*ϕ*(*H*)*∼*=*H*/*H∩N*.

**Theorem 11.13 Correspondence Theorem.** *LetN* *be a normal subgroup of a groupG.*
*ThenH* *7→H*/*N* *is a one-to-one correspondence between the set of subgroups* *H* *containing*
*N* *and the set of subgroups of* *G*/*N. Furthermore, the normal subgroups ofGcontainingN*
*correspond to normal subgroups of* *G*/*N.*

Proof. Let *H* be a subgroup of*G* containing *N*. Since *N* is normal in *H,H*/*N* makes
sense. Let *aN* and *bN* be elements of *H*/*N*. Then (*aN*)(*b−*1*N*) =*ab−*1*N* *∈* *H*/*N*; hence,
*H*/*N* is a subgroup ofG/*N*.

Let *S* be a subgroup of *G*/*N*. This subgroup is a set of cosets of *N*. If *H* = *{g* *∈G*:

*gN* *∈* *S}*, then for *h*1*, h*2 *∈* *H, we have that* (*h*1*N*)(*h*2*N*) = *h*1*h*2*N* *∈* *S* and *h−*11*N* *∈* *S.*

Therefore, *H* must be a subgroup of *G. Clearly,* *H* contains *N*. Therefore, *S* = *H*/*N.*
Consequently, the map*H* *7→H*/*N* is onto.

Suppose that *H*1 and *H*2 are subgroups of *G* containing *N* such that *H*1/*N* = *H*2/*N.*

If *h*1 *∈* *H*1, then *h*1*N* *∈* *H*1/*N*. Hence, *h*1*N* = *h*2*N* *⊂* *H*2 for some *h*2 in *H*2. However,

since*N* is contained in*H*2, we know that *h*1 *∈H*2 or*H*1 *⊂H*2. Similarly, *H*2 *⊂H*1. Since

*H*1=*H*2, the map *H7→H*/*N* is one-to-one.

Suppose that *H* is normal in *G* and *N* is a subgroup of *H. Then it is easy to verify*
that the map *G*/*N* *→G*/*H* defined by *gN* *7→gH* is a homomorphism. The kernel of this
homomorphism is*H*/*N*, which proves that *H*/*N* is normal in*G*/*N.*

Conversely, suppose that*H*/*N* is normal in*G*/*N*. The homomorphism given by
*G→G*/*N* *→* *G*/*N*

*H*/*N*
has kernel*H. Hence,H* must be normal in*G.*

Notice that in the course of the proof of Theorem11.13, we have also proved the following theorem.

**Theorem 11.14 Third Isomorphism Theorem.** *Let* *G* *be a group and* *N* *and* *H* *be*
*normal subgroups of* *G* *with* *N* *⊂H. Then*

*G*/*H* *∼*= *G*/*N*

*H*/*N.*
**Example 11.15.** By the Third Isomorphism Theorem,

Z/*m*Z*∼*= (Z/*mn*Z)/(*m*Z/*mn*Z)*.*
Since *|*Z/*mn*Z*|*=*mn*and *|*Z/*m*Z*|*=*m, we have* *|m*Z/*mn*Z*|*=*n.*

11.3. EXERCISES 129
**Sage** Sage can create homomorphisms between groups, which can be used directly as
functions, and then queried for their kernels and images. So there is great potential for
exploring the many fundamental relationships between groups, normal subgroups, quotient
groups and properties of homomorphisms.

**11.3**

**Exercises**

**1.** Prove that det(*AB*) = det(*A*)det(*B*) for *A, B* *∈* *GL*2(R). This shows that the

determinant is a homomorphism from *GL*2(R) toR*∗*.

**2.** Which of the following maps are homomorphisms? If the map is a homomorphism,
what is the kernel?

(a) *ϕ*:R*∗* *→GL*2(R) defined by
*ϕ*(*a*) =
(
1 0
0 *a*
)
(b) *ϕ*:R*→GL*2(R) defined by
*ϕ*(*a*) =
(
1 0
*a* 1
)
(c) *ϕ*:*GL*2(R)*→*R defined by
*ϕ*
((
*a* *b*
*c* *d*
))
=*a*+*d*
(d) *ϕ*:*GL*2(R)*→*R*∗* defined by
*ϕ*
((
*a* *b*
*c* *d*
))
=*ad−bc*
(e) *ϕ*:M2(R)*→*Rdefined by
*ϕ*
((
*a* *b*
*c* *d*
))
=*b,*

where_{M}2(R) is the additive group of2*×*2 matrices with entries inR.

**3.** Let *A* be an *m×n* matrix. Show that matrix multiplication, *x* *7→* *Ax, defines a*
homomorphism *ϕ*:R*n→*R*m*.

**4.** Let *ϕ*:Z*→*Z be given by*ϕ*(*n*) = 7*n. Prove thatϕ*is a group homomorphism. Find
the kernel and the image of*ϕ.*

**5.** Describe all of the homomorphisms from_{Z}_{24} to_{Z}_{18}.
**6.** Describe all of the homomorphisms fromZtoZ12.

**7.** In the group_{Z}_{24}, let *H*=*⟨*4*⟩* and *N* =*⟨*6*⟩*.

(a) List the elements in *HN* (we usually write *H* +*N* for these additive groups) and
*H∩N*.

(b) List the cosets in *HN*/*N*, showing the elements in each coset.
(c) List the cosets in *H*/(*H∩N*), showing the elements in each coset.

(d) Give the correspondence between *HN*/*N* and *H*/(*H∩N*) described in the proof of
the Second Isomorphism Theorem.

**8.** If *G* is an abelian group and *n* *∈* N, show that *ϕ* : *G* *→* *G* defined by *g* *7→* *gn* is a
group homomorphism.

**9.** If *ϕ* : *G* *→* *H* is a group homomorphism and *G* is abelian, prove that *ϕ*(*G*) is also
abelian.

**10.** If *ϕ* : *G* *→* *H* is a group homomorphism and *G* is cyclic, prove that *ϕ*(*G*) is also
cyclic.

**11.** Show that a homomorphism defined on a cyclic group is completely determined by
its action on the generator of the group.

**12.** If a group*G* has exactly one subgroup*H* of order*k, prove that* *H* is normal in*G.*
**13.** Prove or disprove: _{Q}/Z*∼*=Q.

**14.** Let*G*be a finite group and*N* a normal subgroup of*G. IfH* is a subgroup of*G*/*N,*
prove that*ϕ−*1(*H*)is a subgroup in*G*of order* _{|}H|·|N|*, where

*ϕ*:

*G→G*/

*N*is the canonical homomorphism.

**15.** Let *G*1 and *G*2 be groups, and let *H*1 and *H*2 be normal subgroups of *G*1 and

*G*2 respectively. Let *ϕ* : *G*1 *→* *G*2 be a homomorphism. Show that *ϕ* induces a natural

homomorphism *ϕ*: (*G*1/*H*1)*→*(*G*2/*H*2) if*ϕ*(*H*1)*⊂H*2.

**16.** If*H* and*K* are normal subgroups of*G*and*H∩K*=*{e}*, prove that*G*is isomorphic
to a subgroup of*G*/*H×G*/*K.*

**17.** Let*ϕ*:*G*1 *→G*2be a surjective group homomorphism. Let*H*1be a normal subgroup

of *G*1 and suppose that *ϕ*(*H*1) =*H*2. Prove or disprove that*G*1/*H*1 *∼*=*G*2/*H*2.

**18.** Let *ϕ*:*G→H* be a group homomorphism. Show that*ϕ* is one-to-one if and only if
*ϕ−*1(*e*) =*{e}*.

**19.** Given a homomorphism*ϕ*:*G→H* define a relation *∼*on *G*by *a∼b* if*ϕ*(*a*) =*ϕ*(*b*)

for *a, b* *∈* *G. Show this relation is an equivalence relation and describe the equivalence*
classes.